Exercise 3.2.6

For these problems we can write down the matrix representation of the transformation ${[T]}_{\alpha }^{\beta }$, where $\alpha =\{u_1,u_2,\dots ,u_n\}$ and $\beta =\{v_1,v_2,\dots ,v_n\}$ are the (standard) basis for the domain and the codomain of $T$. And the inverse of this matrix would be $B={[T^{-1}]}_{\beta }^{\alpha }$. So $T^{-1}$ would be the linear transformation such that

1.

We get ${[T]}_{\alpha }^{\beta }=\left(\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right)$ and ${[T^{-1}]}_{\beta }^{\alpha }=\left(\begin{array}{ccc}\hfill -1\hfill & \hfill -2\hfill & \hfill -10\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right)$. So we know that

$$T(a+\mathit{bx}+c{x}^2)=(-a-2b-10c)+(-b-4c)x+(-c)x^2.$$

2.

We get ${[T]}_{\alpha }^{\beta }=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$, a matrix not invertible. So $T$ is not invertible.

3.

We get ${[T]}_{\alpha }^{\beta }=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$ and ${[T^{-1}]}_{\beta }^{\alpha }=\left(\begin{array}{ccc}\hfill \frac{1}{6}\hfill & \hfill -\frac{1}{3}\hfill & \hfill \frac{1}{2}\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 0\hfill & \hfill -\frac{1}{2}\hfill \\ \hfill -\frac{1}{6}\hfill & \hfill \frac{1}{3}\hfill & \hfill \frac{1}{2}\hfill \end{array}\right)$. So we know that

$$T(a,b,c)=(\frac{1}{6}a-\frac{1}{3}b+\frac{1}{2}c,\frac{1}{2}a-\frac{1}{2}c,-\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c).$$

4.

We get ${[T]}_{\alpha }^{\beta }=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ and ${[T^{-1}]}_{\beta }^{\alpha }=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill \frac{1}{2}\hfill & \hfill -\frac{1}{2}\hfill & \hfill 0\hfill \\ \hfill \frac{1}{2}\hfill & \hfill \frac{1}{2}\hfill & \hfill -1\hfill \end{array}\right)$. So we know that

$$T(a,b,c)=(c,\frac{1}{2}a-\frac{1}{2}b,\frac{1}{2}a+\frac{1}{2}b-c).$$

5.

We get ${[T]}_{\alpha }^{\beta }=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right)$ and ${[T^{-1}]}_{\beta }^{\alpha }=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -\frac{1}{2}\hfill & \hfill 0\hfill & \hfill \frac{1}{2}\hfill \\ \hfill \frac{1}{2}\hfill & \hfill -1\hfill & \hfill \frac{1}{2}\hfill \end{array}\right)$. So we know that

$$T(a+\mathit{bx}+c{x}^2)=(b,-\frac{1}{2}a+\frac{1}{2}c,\frac{1}{2}a-b+\frac{1}{2}c).$$

6.

We get ${[T]}_{\alpha }^{\beta }=\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$, a matrix not invertible. So $T$ is not invertible.